draw the following vector quantity Using the coordinate system.
a. 190 newton east
b. 120km/hr, 250 north of east
c. 60 meters southwest​

Answers

Answer 1

The given vectors quantities can be described by their properties of both

magnitude and direction.

a. The drawing of the vector extending from point (0, 0) to (190, 0) on the coordinate plane is attached.b. The velocity vector extending from  (0, 0) to (108.76, 50.714) on the coordinate plane is attached.c. The displacement vector extending from (0, 0) to (30·√2, 30·√2) is attached.

Reasons:

a. The magnitude of the vector = 190 N

The direction in which the vector acts = East

Therefore, in vector form, we have;

[tex]\vec{F}[/tex] = 190 × cos(0)·i + 190 × sin(0)·j = 190·i

The vector can be represented by an horizontal line, 190 units long

Coordinate points on the vector = (0, 0) and (190, 0)

The drawing of the vector with the above points using MS Excel is attached.

b. Magnitude of the velocity vector = 120 km/hr. 25° North of east

Solution;

The vector form of the velocity is; [tex]\vec{v}[/tex] = 120 × cos(25)·i + 120×sin(25)·j, which gives;

[tex]\vec{v}[/tex] = 120 × cos(25)·i + 120×sin(25)·j ≈ 108.76·i + 50.714·j

[tex]\vec{v}[/tex] ≈ 108.76·i + 50.714·j

Therefore, points that define the vector are; (0, 0) and (108.76, 50.714)

The drawing of the vector is attached

c. The magnitude of the vector = 60 m

The direction of the vector is southwest = West 45° south

The vector form of the displacement is [tex]\vec{d}[/tex] = 60 × cos(45°)·i + 60 × sin(45°)·j

Which gives;

[tex]\vec{d}[/tex] = 30·√2·i + 30·√2·j

Points on the vector are therefore; (0, 0), and (30·√2, 30·√2)

The drawing of the vector is attached

Learn more about vectors here:

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Draw The Following Vector Quantity Using The Coordinate System. A. 190 Newton East B. 120km/hr, 250 North
Draw The Following Vector Quantity Using The Coordinate System. A. 190 Newton East B. 120km/hr, 250 North
Draw The Following Vector Quantity Using The Coordinate System. A. 190 Newton East B. 120km/hr, 250 North

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Express the sum of 2.52 kg + 131 g in kilograms with the correct number of significant
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Answers

Convert second term into kg

131 = 0.131

Term has become:

2.52 + 0.131

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Abigail runs one complete lap (400m) around the track, while Gabi runs a 50 meter dash in a straight line. Which runner had a greater displacement?

Answers

Answer:

The distance of one lap on this track from the start line to the finish line is 400 meters. Two laps around the track is 800 meters, half a lap is 200 meters, and so on.

    Remember, displacement is the direction from the starting point and the length of a straight line from the starting point to the ending point. If a runner was going to run 400 meters (one lap around the track), they would start and stop at the same place on the track. Therefore, their displacement would be 0.

 

Explanation:

hope this helped

Answer:

[tex]\boxed {\boxed {\sf Gabi \ had \ greater \ displacement}}[/tex]

Explanation:

Displacement is the change in the position of an object.

Abigail runs a complete lap around the track, which is 400 meters. Even though she ran 400 meters, she has no displacement. If she starts and ends at the same spot, there is no change in position.

Gabi runs a 50 meter dash in a straight line. Gabi has 50 meters of displacement. She runs in a straight line and is 50 meters away from where she began.

While Abigail ran the farther distance, Gabi had the greater displacement.

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The power P in watts (W) is equal to the energy E in joules (J), divided by the time period t in seconds (s):
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Answer:

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Explanation:

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Answer:

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Answers

Answer:

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Explanation:

Answer:

[tex]\huge\boxed{\sf Wavelength= 470\ m}[/tex]

Explanation:

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[tex]\rule[225]{225}{2}[/tex]

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Answers

Hi there!

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Answers

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The pail’s minimum speed at the top of the circle if no water is to spill out is; v = 4.722 m/s

We are given;

Mass of the pail of water; m = 2000 g = 2kg

Radius of the circle; r = 1 m

Tension exerted by string; T = 25 N

To calculate the net force on the pail of water at the top of the vertical circle, let us take equilibrium of forces to give;

W - T = ma

Where;

W is weight = mg

T is tension on the string

Now, in circular motion, we know that;

Acceleration is; a = v²/r

Thus;

mg - T = m(v²/r)

Making v the subject of the formula gives us;

v = √[(r(T + mg)/m)

Plugging in the relevant values gives;

v = √[(1(25 + (2 * 9.8))/2)

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From one point on the ground, the
angle of elevation of the peak of a
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mountain, the angle of elevation is
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一企一
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Answers

Trigonometry allows to find the result for the question about the height of the mountain is:

The height mountain  is:  y = 9674.4 ft

Trigonometry allows finding relationships between the angles of a right triangle.

         [tex]tan \theta = \frac{y}{x}[/tex]  

Where θ is the angle, y the opposite leg (height) and x the adjacent leg (horizontal distance).

In the attachment we can see a diagram of the system. They indicate that for x  distance the angle is 14.67º  

         tan 14.67 = [tex]\frac{y}{x}[/tex]  

At the other point the angle is 10.38º.

        tan 10.38 = [tex]\frac{y}{x+15860}[/tex]

We look for the horizontal distance (x) with these equations.

        x tan 14.67 = (x + 15860) tan 10.38

        x tan 14.67 / tan 10.38 = x + 15860

        x 1,429 = x + 15860

        x 0.429 = 15860

        x = [tex]\frac{15680}{0.429}[/tex]

        x = 36955.3 ft

We calculate for the height.

        y = x tan 14.67

        y = 36955.3 tan 14.67

        y = 9674.4 ft

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